. Final submission March 21, 2010 compiled: March 21, 2010 ON THE n-back-and-forth TYPES OF BOOLEAN ALGEBRAS

. Final submission March 21, 2010 compiled: March 21, 2010 ON THE n-back-and-forth TYPES OF BOOLEAN ALGEBRAS KENNETH HARRIS AND ANTONIO MONTALBÁN Abstract. The objective of this paper is to uncover the structure of the back-andforth equivalence classes at the finite levels for the class of Boolean algebras. As an application, we obtain bounds on the computational complexity of determining the back-and-forth equivalence classes of a Boolean algebra for finite levels. This result has implications for characterizing the relatively intrinsically Σ 0 n relations of Boolean algebras as existential formulas over a finite set of relations. 1. Introduction The program of effective mathematics is concerned with computational aspects of mathematical structures. One asks questions like the following: what is the simplest way to represent a certain structure? Given two isomorphic computable structures, what conditions guarantee that some isomorphism is computable? Or, that every isomorphism is computable? What conditions on a relation in a computable structure guarantee that the image of the relation in any computable copy of the structure is computable? Or, that the image is computably enumerable? How do solutions to these problems generalize within the hyperarithmetic hierarchy? Each of the questions in the previous paragraph has been resolved by providing an effective version of the classical construction of isomorphisms by the back-and-forth method. (See [AK00, Chapters 16-18] for a statement and resolution of each of these questions.) Answering these questions for a specific class of structures requires an analysis of the complexity of the back-and-forth relations on those structures. This paper provides such an analysis of the back-and-forth relations at finite levels for Boolean algebras in order to apply the effective back-and-forth methods for building computable copies of Boolean algebras. The back-and-forth method, or the method of extension of partial isomorphisms, originated with Cantor s stepwise construction of an isomorphism between any two countable dense linear orders without endpoints. The method is basic in model theory, for example, in the proof that any two countably homogeneous models realizing the same types are isomorphic. Langford extended the application of the method to show that any two dense linear orders without endpoints were elementarily equivalent 2000 Mathematics Subject Classification. 03D80. Key words and phrases. Boolean algebra, back-and-forth, invariant. The second author was partially supported by NSF Grants DMS-0600824 and DMS-0901169 and by the Marsden Foundation of New Zealand, via a postdoctoral fellowship. The authors would like to thank Pavel Alaev for comments on a previous draft and the anonymous referee for a detailed report. 1

2 KENNETH HARRIS AND ANTONIO MONTALBÁN (regardless of their cardinality). Ehrenfeucht and Fraïssé gave a purely algebraic characterization of elementary equivalence in first-order structures in terms of families of partial isomorphisms with a (one-at-a-time) back-and-forth property. Karp generalized this characterization and showed that the mathematical framework where the technique is naturally expressed is the infinitary logic L ω. For more details on this history and Karp s contribution, see [Bar73] or [Dic85, Section 4]. The investigation of general back-and-forth techniques in the context of effective mathematics started with the work of Ash on α-systems in [Ash86b] and [Ash86a]. The objects studied are countable structures over a computable language, and the constructions are over computable ordinals. Ash and Knight produced a general metatheorem for applying the α-system machinery. The backbone for applying the metatheorem is a hierarchy of back-and-forth relations on finite approximations of the construction. The construction in the Ash and Knight metatheorem can be viewed as a hierarchy of worker constructions, where the worker at each level has limited information about the structure to be constructed; the back-and-forth relations ensure that mistakes made by workers at lower levels can be repaired at higher levels later in the construction. Success in the construction depends on the workers at each level having access to the appropriate back-and-forth relations to guide their constructions. (See [AK00, Chapters 13 and 14] for more details.) To apply the metatheorem to a particular type of structure, one has to understand how the back-and-forth relations behave for that structure. These relations have been determined for some types of structure. Ash studied the back-and-forth relations on well-orderings [Ash87] and on superatomic Boolean algebras [Ash86b]; Ash and Knight [AK00] studied vector spaces; Barker [Bar95] and Calvert [Cal05] studied reduced Abelian p-groups; and Csima, Montalbán and Shore [CMS06] analyzed saturated Boolean algebras. We are interested in the back-and-forth relations for Boolean algebras at finite levels of the hierarchy of back-and-forth relations. Let L be the language of Boolean algebras and L ω1 ω the infinitary language allowing countable conjunctions and disjunctions, but finite nesting of quantifiers. The subclasses of infinitary formulas Σ n and Π n have n alternations of existential and universal quantifiers (counting infinite disjunctions as existential quantification and infinite conjunctions as universal quantification). We write Σ n (A) (Π n (A)) for the set of Σ n (Π n ) sentences true in A. The n-back-and-forth relation between two Boolean algebras holds when Π n (A) Π n (B) (or, equivalently, when Σ n (B) Σ n (A), see Subsection 2.3). We write A n B when either of these two conditions is obtained and A n B when both A n B and B n A. Our first aim in this paper is to provide a combinatorial characterization of the n- back-and-forth equivalence classes. We aim for a map T n ( ) on Boolean algebras so that A n B T n (A) = T n (B); furthermore, we want the object given by T n (A) to be computationally simpler than Σ n (A), so that determining whether T n (A) = T n (B) is, ideally, computable. One of the main results of this paper is to produce invariants which satisfy these two conditions (Section 7). The definition of these invariants is based on a modification of a topological

ON THE n-back-and-forth TYPES OF BOOLEAN ALGEBRAS 3 invariant defined by Flum and Ziegler [FZ80]. To each countable Boolean algebra A, we assign an object, T n (A), which encodes how A may be partitioned into a finite join of elements. Let INV n be the set of possible values of T n (A) (we will show that INV n is countable). On the elements of INV n, we define a partial ordering n, a binary operation +, and a function ( ) n 1 : INV n INV n 1 satisfying the following properties for all Boolean algebras A and B: T n (A) n T n (B) A n B, T n (A) + T n (B) = T n (A B), (T n (A)) n 1 = T n 1 (A) (where A B is the directed sum of A and B). The definitions are purely combinatorial, and the structures (given for each n) (INV n, n, +, ( ) n ) are uniformly computable in n. The key to our investigation of the n-back-and-forth types are n-indecomposable Boolean algebras: Definition 3.1. A Boolean algebra A is n-indecomposable if for any partition of A into subalgebras, A = A 0... A k, there is an i k such that A n A i. We summarize the main results about n-indecomposable algebras. (a) For each n, there are only finitely many n-back-and-forth equivalence classes among the n-indecomposable algebras (Theorem 3.17). (b) Every Boolean algebra can be decomposed into a finite sum of n-indecomposable subalgebras (Theorem 3.13). (c) These decompositions fully determine the n-back-and-forth equivalence classes among all Boolean algebras (Lemma 3.14). BF n will denote the set of invariants in INV n which correspond to n-indecomposable Boolean algebras. By (b) and (c), INV n is finitely generated by BF n under +. The general n-back-and-forth invariants, INV n, record all the possible ways an algebra can be partitioned into finitely many n-indecomposable subalgebras, up to n-back-and-forth equivalence. A second aim of this paper is to provide an analysis of the computational complexity of the relations on a Boolean algebra A determined by T n (A a) n σ for a A and σ INV n. We use, for this purpose, the hierarchy of computable infinitary formulas, Σ c n and Π c n, which restricts conjunctions and disjunctions to computably enumerable sets of formulas (see Subsection 2.2). The important property of this class is that for any formula φ(x) Π c n, the relation {a A : A = φ(a)} is Π 0 n. We show that for each σ INV n, there is a Π c n formula φ c σ(x) and a Π c n+1 formula ψ c σ(x) such that for each a A, (Lemmas 8.10 and 8.11). A = φ c σ(a) T n (A a) n σ A = ψ c σ(a) T n (A a) n σ

4 KENNETH HARRIS AND ANTONIO MONTALBÁN One application of these invariants is a quantifier elimination result for Σ c n+1 formulas. Let R σ be the unary predicate which holds of an element a A if and only if T n (A a) n σ; then, from the previous paragraph, the set {a A : A = R σ (a)} is a Π 0 n(a) subset of A for any σ INV n. We show that every Σ c n+1 formula in the language of Boolean algebras is equivalent to a Σ 0 1 formula over the finitely many additional predicates R α, for α BF n. Theorem 8.12. Let B be a Boolean algebra, R B and n ω. The following are equivalent. (1) R is relatively intrinsically Σ n+1. That is, if A = B and (A, Q) = (B, R), then Q is a Σ 0 n+1(a) subset of A. (2) R is explicitly Σ n+1. That is, R can be defined in B by a computable infinitary Σ c n+1 formula. (3) There is a 0 (n) -computable sequence {ϕ i : i ω} of finitary Σ 1 formulas over the predicates R α, for α BF n, such that x R i ω ϕ i (x). The equivalence between the first two statements is due to Ash, Knight, Manasse, Slaman; Chisholm (see [AK00, Theorem 10.1]). This theorem says that the finitely many predicates, R α for α BF n, essentially give all the structural information about a Boolean algebra that can be computed in n Turing jumps. The bound on the complexity of the sequence {ϕ i : i ω} cannot be improved from 0 (n) because the Σ c n+1 formulas are strong enough to code any Σ 0 n+1 set (see Section 2.2 for further details). We also have a characterization of when the Σ c n+1-diagram of a Boolean algebra A (that is, the Σ c n+1 sentences with constants from A which are true in A) is Σ 0 n+1. Theorem 8.13. Let A be a presentation of a Boolean algebra. The following are equivalent. (a) The Σ c n+1-diagram of A is a Σ 0 n+1 set of formulas. (b) The relations R α (A) are computable in 0 (n) for each α BF n. A Boolean algebra is n-approximable if it satisfies either of the conditions in the theorem. Note that being n-approximable is a property of a presentation of a Boolean algebra rather than a property of the isomorphism type of a Boolean algebra. Of course, every computable Boolean algebra is n-approximable; and every low n Boolean algebra is also n-approximable. The main open question is whether every n-approximable Boolean algebra has a computable copy. This is true for n {1, 2, 3, 4}: Downey and Jockusch [DJ94] proved that every low Boolean algebra has a computable copy; this was extended by Thurber [Thu95] to low 2 Boolean algebras, and by Knight and Stob [KS00] to low 4. Each proof proceeds by showing that any n-approximable Boolean algebra has an (n 1)-approximable copy, for n {1, 2, 3, 4}. It is still open whether every 5-approximable Boolean algebra has a computable copy. It is sufficient here to show that every 5-approximable Boolean algebra has a 4-approximable copy. A motivation for this work is to obtain a better understanding of the presentations of n-approximable algebras in order to provide a means of determining whether every low n Boolean algebra

ON THE n-back-and-forth TYPES OF BOOLEAN ALGEBRAS 5 has a computable copy. To this end, we have enumerated BF n for n {1, 2, 3, 4, 5} in Section 6. We mention two antecedents of our work. Jockusch and Soare [JS94] produced finitely many invariants for the back-and-forth relations on Boolean algebras through level two, as well as topological invariants for the Stone space of a Boolean algebra along the same lines as Flum and Ziegler [FZ80]. Pavel Alaev [Ala04] described the back-and-forth relations on Boolean algebras through level four by means of numerical invariants and used these invariants in combination with an extension of the Ash-Knight α-system machinery. 2. Preliminaries 2.1. Boolean algebras. Boolean algebras will be taken to be countable and in the signature,,, 0, 1, but otherwise we follow the standard reference [Mon89]. Boolean algebras will be denoted by A, B, C and their elements by a, b, c. We abbreviate a b for a b = a and denote the relative algebra by A a = {b A : b a}. A partition of an element a in a Boolean algebra A is a finite sequence a 0,..., a k of pairwise disjoint elements (that is, a i a j = 0 when i j) such that a = a 0... a k ; a partition of a Boolean algebra A is a partition of its unit, 1 A. We write a = (a i ) i k to mean that a 0,..., a k is a partition of a. If (a i ) i k is any sequence of elements from A, then a sequence (b i,j ) i k,j li is a refinement of (a i ) i k if a i = j l i b i,j. If (a i ) i k and (b j ) j l are partitions, then there is a common refinement partition (c i,j ) i k,j l (where c i,j = a i b j ) satisfying a i = j l c i,j and b j = i k c i,j. Note that any finite sequence of elements (a i ) i k can be refined to a partition (c j ) j 2 k, where each c j is of the form i k b i and each b i is a i or a i. An ultrafilter U of a Boolean algebra A is a nonempty, proper subset of A satisfying (Uf.a) For any partition (a i ) i k of A, a i U for some i k. (Uf.b) For all a U, if b a, then b U. (Uf.c) For all a, b U, a b U. For any a U, we write U a for the set {b U : b a}. If U is an ultrafilter on A, then U a is an ultrafilter on A a. Conversely, if U is an ultrafilter on A a, then there is a unique ultrafilter U on A with U U; it is given by U = { x A : x a U }. The set of ultrafilters of A is denoted by Ult(A). There is a natural topology on Ult(A), the Stone space of A, determined by the basic open sets O a = {U Ult(A) : a U} for a A. This topological space is Hausdorff, compact and zero-dimensional see [Mon89, 7]. 2.2. Syntax. We will be working in a predicate language appropriate for countable Boolean algebras, L ω1 ω, which allows conjunctions and disjunctions over arbitrary countable sets of formulas and finite nesting of quantifiers. The non-logical signature includes the Boolean operators,,, and constants 0, 1; the logical symbols are the connectives (countable conjunction), (countable disjunction), (negation), the quantifiers and, and = (identity). We will also write a = a 0... a k to emphasize that a 0,..., a k is a partition of a. All formulas have only finitely many distinct free variables.

6 KENNETH HARRIS AND ANTONIO MONTALBÁN Satisfaction of formulas of L ω1 ω in a Boolean algebra is defined in the natural way, and A = φ(a 0,..., a k ) means A satisfies the formula φ(x 0,..., x k ) at the tuple of elements (a 0,..., a k ). The subclass of formulas in normal form will be denoted by Σ n and Π n and defined inductively for n ω. The Σ 0 and Π 0 formulas are the finitary open formulas (that is, quantifier-free formulas with no infinite disjunctions or conjunctions). For n > 0, a Σ n formula is a countable disjunction of formulas of the form xψ, where x is a finite sequence of variables and ψ Π m for some m < n. Similarly, a Π n formula is a countable conjunction of formulas of the form xψ, where x is a finite sequence of variables and ψ Σ m for some m < n. The negation of a Σ n formula is logically equivalent to a Π n formula (see [AK00, Lemma 6.1]). The class of computable infinitary formulas, L c ω 1 ω, was introduced in [Ash86b] and restricts conjunctions and disjunctions to computably enumerable sets of formulas. We provide an informal description of the class Σ c n and Π c n for n ω, although the definitions can be extended over the computable ordinals. The Σ c 0 and Π c 0 formulas are the finitary open formulas. The computable infinitary Σ c n+1 formulas are restricted to disjunctions over a c.e. set of formulas of the form xψ, where x is a finite sequence of variables and ψ Π c m for some m n. The computable infinitary Π c n+1 formulas are restricted to conjunctions over a c.e. set of formulas of the form xψ, where x is a finite sequence of variables and ψ Σ c m for some m n. The important fact about computable infinitary formulas is that the complexity of the formula matches the complexity of its interpretation. If ϕ(x 0,..., x k ) Σ c n, then the set of tuples (a 0,..., a k ) for which A = ϕ(a 0,..., a k ) is Σ 0 n(a), and if ϕ(x 0,..., x k ) Π c n, then the set of tuples (a 0,..., a k ) for which A = ϕ(a 0,..., a k ) is Π 0 n(a). See [AK00, Chapter 7] for a formal definition of the computable infinitary formulas and their properties. Our use of the notation Σ c n and Π c n was used in [AK90], although not in [AK00]. The sentences in Σ c n+1 are strong enough to code any Σ 0 n+1 set in the following sense. For any Σ 0 n+1 set X = We 0(n) there is a computable sequence of Σ c n+1 sentences θ e,k k ω such that k X implies θ k and k X implies θ k (see [AK00, Theorem 7.9]). Let ψ k be a Σ c 2 sentence stating there are exactly k elements in the universe; that is, x 1... x k y [ y = x i ] x i x j. 1 i k 1 i<j k For n 1, the sentence θ e,k ψ k is Σ c n+1. By dovetailing computable disjunctions, the sentence ϕ e := k ω (θ e,k ψ k ) can be rewritten equivalently as a Σ c n+1 sentence. Now, A = ϕ e if and only if A = k for some k X. 2.3. Back-and-forth relations. We refer the reader to [AK00, Chapter 15, especially 15.3.4] for more details on back-and-forth relations in Boolean algebras. The back-and-forth relation on Boolean algebras, A n B, was defined in the introduction as Π n (A) Π n (B), or equivalently Σ n (B) Σ n (A). However, it is more convenient to use an alternative characterization. The two accounts of n converge in Theorem 2.3. Definition 2.1. The relation n on Boolean algebras is defined by recursion on n ω. Let A and B be Boolean algebras. Define A 0 B if either both A and B are the one-element Boolean algebra or neither is; define A n+1 B if for every partition (b i ) i k

ON THE n-back-and-forth TYPES OF BOOLEAN ALGEBRAS 7 of B, there is a partition (a i ) i k of A such that B b i n A a i for all i k. We write A n B when both A n B and B n A. For a A and b B define a n b if A a n B b. For any partition (a i ) i k of A and (b i ) i k of B, we write (a i ) i k n (b i ) i k if a i n b i for each i k. So, A n+1 B if and only if for every partition (b i ) i k of B, there is a partition (a i ) i k of A with (b i ) i k n (a i ) i k. The following is essentially from [AK00, Lemma 15.12]. Lemma 2.2. Let (a i ) i k be a partition of A and (b i ) i k be a partition of B satisfying (a) i k n (b) i k. Then A n B. Proof. The proof is by induction on n ω. The result is trivial for the basis case n = 0. Suppose the statement of the lemma is true for n. Let (a i ) i k be a partition of A and (b i ) i k be a partition of B satisfying (a i ) i k n+1 (b i ) i k. To show A n+1 B, consider any partition (d j ) j l of B, and let (b i,j ) i k,j l be the common refinement with (b i ) i k, so that d j = i k b i,j and b i = j l b i,j. Then, for each i k there is a partition (a i,j ) j l of a i where a i,j n b i,j. Let c j = i k a i,j, so that by the inductive hypothesis c j n d j. Thus, (c j ) j l is a partition of A with (c j ) j l n (d j ) j l. Since (d j ) j l was arbitrary, we have A n+1 B by Definition 2.1. The next theorem connects the syntactic characterization of back-and-forth relations from the introduction with the characterization by partitions (see [AK00, Proposition 15.1]). Theorem 2.3. Let (a i ) i k be a partition of A and (b i ) i k be a partition of B. Then for any n ω the following are equivalent. (1) (a i ) i k n (b i ) i k. (2) The Σ n formulas true of (b i ) i k in B are true of (a i ) i k in A. (3) The Π n formulas true of (a i ) i k in A are true of (b i ) i k in B. Since Σ n, Π n Σ n+1, we have the following. Corollary 2.4. If A n+1 B, then A n B. 3. Indecomposable algebras Indecomposable Boolean algebras are the essence of our analysis of back-and-forth invariants. Heuristically, an indecomposable algebra is one which is similar to one of its subalgebras regardless of how it is split, where similarity captures some equivalence relation on Boolean algebras of interest. For example, when the relation is that of isomorphism, a pseudo-indecomposable Boolean algebra A is one satisfying A = A a or A = A a for each a A. In this paper we are interested in the family of back-and-forth equivalence relations n for n ω. Definition 3.1. A Boolean algebra A is n-indecomposable if for any partition (a i ) i k of A, there is an i with A n A a i. We will show in Theorem 4.1 that this definition is equivalent to the condition that A n A a or A n A a for any a A. We will also say more at the end of Section 7 about the relation between pseudo-indecomposable algebras and the notion of indecomposable captured in this definition.

8 KENNETH HARRIS AND ANTONIO MONTALBÁN It is more convenient to approach the characterization of n-indecomposable algebras from a different viewpoint. We start with the topological characterization of the n-types of accumulation points in the Stone space of the algebra and impose an additional relation, n, on these n-types. It follows from the work of Flum and Ziegler [FZ80] that for every ultrafilter U of a Boolean algebra A, there is an a U (the basic neighborhood O a of U in the Stone space) such that for every b U with b a (equivalently in the Stone space, O b O a ), the Boolean algebras A b are all n -equivalent. (See [Hei81] or [FZ80, Part II, 1, C] for further discussion of the classification of the accumulation points in a Stone space. Knowledge of this material is not presupposed in what follows.) We classify accumulation points in the Stone space using equivalence classes under n, the n-backand-forth types; we use these types to pick-out the n-indecomposable algebras (see Definition 3.5). The key theorems are that every Boolean algebra can be decomposed into finitely many n-indecomposable types (Theorem 3.13) and that the n-back-andforth types are invariants for n-indecomposable algebras (Theorem 3.17). These two theorems allow us to describe invariants for the n-back-and-forth equivalence classes of all Boolean algebras by decomposing algebras into finitely many n-indecomposable subalgebras (see Section 7). The heuristic approach to n-indecomposability from Definition 3.1 and our more technically involved definition in terms of accumulation points (Definition 3.5) converge in Theorem 4.1. In Section 6 we provide a description of all n-back-and-forth types for n 5. 3.1. Indecomposable types. In Definition 3.6, we will define finite partial orderings (ABF n, n ) where the elements of ABF n+1 are subsets of ABF n. The reader could jump ahead and read Definition 3.6 now, though it might look too technical at first. We call the elements of ABF n n-bf-types. Before actually defining these partial orderings, we define invariant maps t n : Ult(A) ABF n and commence to assign n-bf-types to Boolean algebras. Definition 3.2. For X ABF n, we let max X be the antichain of n -maximal elements of X. Definition 3.3. We assume A is not the trivial one element algebra. To each ultrafilter U of a Boolean algebra A and each n ω, we assign an n-bf-type as follows. t 0 (U) =, where is a new symbol. ˆt n+1 (U) = { α ABF n : a U V Ult A [ V U & a V & t n (V ) = α ]}. t n+1 (U) = max ˆt n+1 (U). The definition is usefully rephrased in topological terms on the Stone space: ˆt n+1 (U) consists of the elements α ABF n such that U is an accumulation point of {V Ult(A) : t n (V ) = α}. (In the Stone space, the elements a U correspond to basic clopen neighborhoods of U, U O a ; so α ˆt n+1 (U) if each basic clopen neighborhood of U meets {V Ult(A) : t n (V ) = α} in a point distinct from U.) To each nonzero element a A, we assign an n-bf-type as follows. For n = 0, let t 0 (a) =. Given t n, let ˆt n+1 (a) ABF n be the set of α ABF n such that there are infinitely many distinct ultrafilters V with a V and t n (V ) = α; then, let t n+1 (a) = max ˆt n+1 (a). (In the Stone space, α ˆt n+1 (a) if the clopen neighborhood O a contains infinitely many points of {V Ult(A) : t n (V ) = α}.)

ON THE n-back-and-forth TYPES OF BOOLEAN ALGEBRAS 9 We extend the map to Boolean algebras A by t n (A) = t n (1 A ). The invariants defined in [FZ80] just iterate the function ˆt n, and do not mention the antichain of maximal elements; nor do they consider any ordering on the set of invariants. The reason we need to use this refined version of their invariants is that we are seeking an invariant that corresponds to the hierarchy of n-back-and-forth equivalence relations. Both our n-invariants and Flum and Ziegler s record information on how a Boolean algebra can be partitioned into (n 1)-invariants. Our invariants record exactly the amount of information necessary to decide the n-back-and-forth relations (Definition 2.1). Definition 3.4. For X ABF n, we let dc X ABF n be the n -downward closure of X. Definition 3.5. For n > 0 an element a A is n-indecomposable for ultrafilter U a if (i) t n (U) = t n (a) and (ii) for all V U with a V, t n 1 (V ) dc t n (a). For n = 0, every nonzero a A is 0-indecomposable for any ultrafilter U a. A Boolean algebra A is n-indecomposable for ultrafilter U A if 1 A is n-indecomposable for ultrafilter U. We will show in the next section that the following are equivalent for a Boolean algebra A: A is n-indecomposable for the ultrafilter U; A n A a for all a U; and, for every partition (a i ) i k of A there is some i k with A n A a i. In this section, when we write that A is n-indecomposable, we mean that it is n- indecomposable for some ultrafilter U. We now define the abstract n-bf-types, ABF n (n ω). The model for the relation n on ABF n is Definition 2.1 of the n-back-and-forth relation on Boolean algebras and its consequence Corollary 2.4. Definition 3.6. By recursion on n, we define a finite set ABF n (for n > 0, a collection of subsets of ABF n 1 ), a relation n on the subsets of ABF n 1 (so a relation on ABF n by restriction), and a map ( ) n 1 : ABF n ABF n 1. Let ABF 0 = { } and 0. For α ABF n, let (α) n = max{(γ) n 1 : γ α}, if n > 0; and let (α) 0 =. For α, β ABF n, let α n+1 β if (α) n n (β) n and δ β γ α (γ n δ). For α, β ABF n, let α n+1 β if α n+1 β and β n+1 α. Let ABF n+1 be the set of n -antichains of ABF n. Not all these n-bf-types can be realized in a Boolean algebra (hence the name abstract n-bf-types). We will provide conditions that describe which of the n-bf-types are realized in some Boolean algebra in Section 5. Later on, we will use BF n to denote only the set of n-bf-types which are realized as t n (A) for some n-indecomposable A. For now, we work with ABF n, the set of all abstract n-bf-types, realizable or not. We introduce the following convenient notation. Notation 3.7. Let n 1, α, β ABF n, and γ ABF n 1. We will write γ w α when γ dc α (that is, there is a δ α with δ n 1 γ). We will write α w n β when dc β dc α (that is, for every δ β, there is a γ α with δ n 1 γ). We will write α < w n β if α w n β but not β w n α.

10 KENNETH HARRIS AND ANTONIO MONTALBÁN This notation will be useful because given a Boolean algebra A that is n-indecomposable for ultrafilter U and V Ult(A) with V U, both t n 1 (V ) w t n (A) and t n (A) w n t n (V ) hold. (This is a consequence of Lemma 3.10.) Note that if α, β ABF n are n -antichains and α n+1 β, then α = β. The reason is that if γ α, then there is a δ β and ξ α so that so that γ n δ n ξ. Thus, γ = ξ = δ by the maximality of γ, and so α β. Similarly, β α. As in the definitions above, in the next few lemmas, it could be useful for the reader to keep in mind the Stone space Ult(A) of the Boolean algebra being considered. For example, if U Ult(A) and a, b A, one could read a U as U O a, and a b as O a O b. 3.2. Proof of compactness. This subsection and the next are dedicated to showing that n, t n ( ), ( ) n 1, and n-indecomposables have the properties claimed. The proofs are mostly combinatorial and not very hard. We recommend the reader to read the statements and comments, and skip the proofs in a first read of the paper. Lemma 3.8. Let n 0 and α, β ABF n+1. (a) α n+1 β if and only if α w n+1 β and (α) n n (β) n. (b) α w n+1 β implies (α) n w n (β) n. (c) α n+1 β if and only if α w n+1 β and β w n+1 α. (d) α n+1 β if and only if (α) n w n (β) n and α w n+1 β. It follows from (c) that every α ABF n is n+1 -equivalent to the antichain of its n -maximal elements. Proof. (a). This follows from the definition of n+1. (b). Pick δ (β) n and let δ β be such that (δ ) n 1 = δ. By hypothesis there is a γ α with δ n γ. So, δ = (δ ) n 1 n 1 (γ) n 1 w (α) n. Since δ was arbitrary, (α) n w n (β) n. (c). The direction ( ) follows from (a). We now show ( ). Assume for α, β ABF n that β w n+1 α and α w n+1 β. By (b) we have that (α) n = (β) n. Then, by (a), we have that β n+1 α and α n+1 β as wanted. (d). The direction ( ) follows from (a). For the other direction, by (b), (α) n w n (β) n and then by (c) and hypothesis, (α) n n (β) n. Then, use (a) again. The following facts are easily verified, and will be appealed to without comment: a b implies t n (b) w n t n (a). a = a 0... a k implies t n (a) n max i t n (a i ). The assignments t n respect the operation ( ) n 1. Lemma 3.9. Let A be any Boolean algebra. ( (1) t n (U) n tn+1 (U) ), for every U Ult A. ( n (2) t n (a) n tn+1 (a) ), for every a A. n Proof. (1): The proof is by induction, where ( the basis case n = 0 is trivial. Assume it is true for n 1. We show (i) t n (U) w n tn+1 (U) ) and (ii)( t n n+1 (U) ) n w n t n (U). The conclusion follows by Lemma 3.8.c. For (i), suppose that α t n+1 (U) with ( α ) ( t n 1 n+1 (U) ). Then for any a n U there is an ultrafilter V a distinct from U with t n (V ) = α. By the inductive

ON THE n-back-and-forth TYPES OF BOOLEAN ALGEBRAS 11 ( hypothesis, t n 1 (V ) n 1 tn (V ) ) = (α) n 1 n 1, and hence (α) n 1 ˆt n (U). So, by the definition ( of t n (U), there is a γ t n (U) with (α) n 1 n 1 γ. We conclude that t n (U) w n tn+1 (U) ). n For (ii), we will prove the stronger condition that t n (U) ( t n+1 (U) ). Let α t n n(u) and suppose for reductio that there is no β t n+1 (U) with (β) n 1 n 1 α; that is, for each β ABF n with (β) n 1 n 1 α, there is a β U such that no V a β distinct from U satisfies t n (V ) = β. Since ABF n is finite, list the n-bf-types with (β) n 1 n 1 α as {β 0,..., β l }, and choose witnesses {a β0,..., a βl } as by the reductio supposition. Let a = i l a β i, so that a U, and let V a distinct from U with t n 1 (V ) = α. Then by the inductive hypothesis, ( t n (V ) ) = α, so that t n 1 n(v ) = β i (for some i l). But then a βi V contradicting the choice of a βi. Thus, for some β ABF n with (β) n 1 n 1 α we have β t n+1 (U). ( (2): We will show (i) t n (a) w n tn+1 (a) ) and (ii)( t n n+1 (a) ) n w n t n (a). The conclusion follows by Lemma 3.8.c. For (i), suppose α t n+1 (a) with (α) n 1 ( t n+1 (a) ). Then there are infinitely many n ( ultrafilters V a with t n (V ) = α. By (1), for each such V, t n 1 (V ) n 1 tn (V ) ), n 1 so that for some γ t n (a), (α) n 1 n 1 γ. For (ii), suppose α t n (a). Since ABF n is finite, there is a β ABF n with infinitely many ultrafilters V a satisfying t n (V ) = β and (β) n 1 n 1 α. So, there is a γ t n+1 (a) with β n γ, and thus (γ) n 1 n 1 (β) n 1 n 1 α. The next three lemmas are dedicated to proving that for every ultrafilter U of A and for every n, there exists a U such that a is n-indecomposable for U (Lemma 3.12.1). Lemma 3.10. Let n > 0. Then for every ultrafilter U and every a U, t n (a) w n t n (U). In particular, if A is n-indecomposable for ultrafilter U and V Ult(A) is distinct from U, then t n 1 (V ) w t n (A) and t n (A) w n t n (V ). Note that a U does not imply that t n (a) = t n (U). Proof. Fix a, an ultrafilter U a, and let α t n (U). We argue by induction that for each k ω there are at least k ultrafilters containing a whose (n 1)-bf-type is α and are each distinct from U. Given k such ultrafilters V 0,..., V k 1, there is a b < a with b U V i for all i < k. By the definition of t n (U), there is an ultrafilter V k distinct from U with b V k and t n 1 (V k ) = α. Thus, we have a V k, and V k is distinct from U and each V i (i < k). This completes the induction. From the definition of t n (a), there is a γ t n (a) such that α n 1 γ. For the second part of the lemma, note that t n 1 (V ) w t n (A) by Definition 3.5 and t n (A) w n t n (V ) by the first part of the lemma, since 1 A V. Lemma 3.11. For every ultrafilter U, there is an a U such that t n (U) = t n (a). Proof. Let U be an ultrafilter for the Boolean algebra A. We show that if a U and t n (a) < w n t n (U), then there is a b U with b < a and t n (a) < w n t n (b) w n t n (U). The lemma follows as t n (1 A ) w n t n (U) (by Lemma 3.10) and t n (1 A ) is a finite set. Suppose a U and t n (a) < w n t n (U); let α t n (a) but no γ t n (U) satisfies α n 1 γ. For each γ n 1 α there is a c γ U for which there is no ultrafilter V c γ distinct from U with t n 1 (V ) = γ. Let I = {γ : γ n 1 α} and b = a γ I c γ. So, b U and for no γ n 1 α is it the case that γ t n (b). Thus, t n (a) < w n t n (b) w n t n (U).

12 KENNETH HARRIS AND ANTONIO MONTALBÁN Lemma 3.12. (1) For every ultrafilter U, there is an a U which is n-indecomposable for U. (2) If a is n-indecomposable for an ultrafilter U, then every b U a is also n- indecomposable for U. Proof. (1): Let U be an ultrafilter. By Lemma 3.11 there exists b U with t n (b) = t n (U). For each α ABF n 1, let k α be the number of ultrafilters (possibly infinite) V U with b V and α = t n 1 (V ). Since t n (b) = t n (U), for every α w t n (U), k α is finite. Let U = {V Ult(A) : b V & V U & t n 1 (V ) w t n (U)}. The set U is finite. So, there exists a b with a U but a V for any V U. Since t n (b) w n t n (a) w n t n (U) = t n (b), we have that t n (a) = t n (U). Also, for any V with a V, since V U, t n 1 (V ) w t n (U). It follows that a is n-indecomposable for U. (2): Let a be n-indecomposable for ultrafilter U and b U a. We show that b satisfies conditions (i) and (ii) of Definition 3.5 for U. Since b a, t n (U) = t n (a) w n t n (b) w n t n (U), so t n (U) = t n (b), and (i) holds. Since (ii) holds of a, b a and t n (a) = t n (b), (ii) holds of b. From now on we will only apply the map t n to n-indecomposable algebras. The following compactness theorem follows from the compactness of Ult(A), and plays a key role in the development that follows. Theorem 3.13. For every n ω and every Boolean algebra A, there is a partition (a i ) i k of A where each a i is n-indecomposable. Proof. For each ultrafilter U, fix a U U which is n-indecomposable for U. Since Ult(A) is a compact topological space, the open covering {O au : U Ult(A)} has a finite subcovering. We include a proof of this: Suppose, toward a contradiction, that for every finite set {a U0,..., a Uk }, we have 1 A a U0... a Uk. So 0 A a U0... a Uk. Thus, there is a (proper) ultrafilter V extending the set { a U : U Ult A } (see [Mon89, Proposition 2.16]). But, in this case, a V, a V V which contradicts V is proper. Thus, there are n-indecomposable elements a U0,..., a Ul for filters U 0,...,U l such that 1 A = a U0... a Ul. Using Lemma 3.12.2 we may refine (a Ui ) i l to a partition of A as required. 3.3. Proof of invariance. Our goal now is to show that for n-indecomposable Boolean algebras A and B, A n B if and only if t n (A) n t n (B). The following lemma refines the definition of the n-back-and-forth relations. Lemma 3.14. For all Boolean algebras A and B and for every n ω, A n+1 B if and only if for every partition (b i ) i k of B into n-indecomposables, there is a partition (a i ) i k of A such that (b i ) i k n (a i ) i k. Note that there is no claim that the partition of A given by the theorem is into n-indecomposables. Indeed, this may not be possible (see example 7.21). Proof. ( ) follows from Definition 2.1. For ( ): given any partition (b i ) i k, we can refine this partition into a partition of n-indecomposables using Theorem 3.13 on B b i ; then, apply Lemma 2.2 to the matching partition from A.

ON THE n-back-and-forth TYPES OF BOOLEAN ALGEBRAS 13 Lemma 3.15. Let a be (n + 1)-indecomposable for ultrafilter U and α ABF n. (1) If α w t n+1 (a), then for any k ω there is a partition (a i ) i k of a with a 0 U and α n t n (a i ) (for 1 i k), where each a i is n-indecomposable for some ultrafilter. (2) If (a i ) i k is a partition of a with a 0 U and each a i (1 i k) is n- indecomposable for some ultrafilter, then t n (a i ) w t n+1 (a) for each 1 i k. Proof. (1). Suppose that α w t n+1 (a), so that for some γ ABF n, α n γ t n+1 (a). Then there are distinct ultrafilters V 1,..., V k, all distinct from U, with a V i and γ = t n (a i ) for i k (see proof of 3.10). Fix a 1,..., a k < a such a i is n-indecomposable for V i but a i U, V j for any j i. Moreover, choose the a i to be pairwise disjoint. Let a 0 = a (a 1... a k ), then a 0 U. (2). Let a = a 0... a k where a 0 U and each a i (1 i k) is n-indecomposable for some ultrafilter V i. Then V i U and a V i, so that t n (V i ) w t n+1 (a) by condition (ii) of Definition 3.5. Thus, t n (a i ) w t n+1 (a), for i 1. The following gives one-half of a back-and-forth characterization of when n-indecomposable algebras have the same n-bf-type (the converse follows by Theorem 3.17 and Lemma 4.3). Lemma 3.16. Let n > 0 and A and B be n-indecomposable for ultrafilters U and V, respectively. If t n (B) n t n (A) then for every partition (a i ) i k of A into (n 1)- indecomposable elements, there is a partition (b i ) i k of B into (n 1)-indecomposable elements with t n 1 (a i ) n 1 t n 1 (b i ) for each i k. Proof. Suppose that t n (B) n t n (A). Note that t n (V ) n t n (U) and t n (1 B ) n t n (1 A ). The proof is by induction on the length of the partition k, where k = 0 holds by supposition. Suppose for induction that for any partition (c i ) i k 1 of A where c 0 U and each c i is (n 1)-indecomposable, there is a partition (d i ) i k 1 of B where d 0 V, each d i is (n 1)-indecomposable, and such that t n 1 (c i ) n 1 t n 1 (d i ) for each i < k. Consider any partition (a i ) i k of A, where a 0 U and each a i is (n 1)-indecomposable. Then, a 0 a k U and is n-indecomposable for U (since 1 A is n-indecomposable for U by Lemma 3.12.2). By the inductive hypothesis there is a matching partition of B, (b i ) i k 1, where b 0 V (so n-indecomposable by Lemma 3.12.2), each b i is (n 1)- indecomposable, and such that t n 1 (a i ) n 1 t n 1 (b i ) for all i < k. Since b 0 V and (a 0 a k ) U and both are n-indecomposable, it follows that t n (b 0 ) n t n (a 0 a k ). As a k is (n 1)-indecomposable, let U k be an ultrafilter for which for which it is (n 1)- indecomposable; and let α = t n 1 (a k ) = t n 1 (U k ). Since a 0 a k is n-indecomposable, there exists some γ t n (a 0 a k ) with α n 1 γ. But then there is a δ t n (b 0 ) with γ n 1 δ, so that there is an ultrafilter V k b 0 distinct from V with t n 1 (V k ) = δ. Partition b 0 as b 0 b k where b k is (n 1)-indecomposable for V k and b 0 V. Thus, we have t n 1 (a k ) n 1 t n 1 (b k ) and t n 1 (a 0 ) n 1 t n 1 (b 0). We arrive at the main result of this section.

14 KENNETH HARRIS AND ANTONIO MONTALBÁN Theorem 3.17. Let A and B be n-indecomposable Boolean algebras. Then A n B t n (A) n t n (B). Proof. The proof is by induction on n. We will simultaneously prove the following result, which will be of independent interest. Lemma 3.18. Let a A be n-indecomposable and d 0,..., d k B satisfy a n i k d i. Then a n d i for some i k. When n = 0 the lemma and the theorem hold trivially. We assume the theorem holds for n 1, and will later show that this implies that the lemma holds for n. First, we use that the lemma holds for n to prove the theorem for n. ( ). Suppose t n (A) n t n (B). To prove A n B, consider any partition of B, (b i ) i k, into (n 1)-indecomposables. (This is sufficient by Lemma 3.14.) By Lemma 3.16 there is a matching partition of A, (a i ) i k, into (n 1)-indecomposables such that t n 1 (b i ) n 1 t n 1 (a i ) (for each i k). By the inductive hypothesis, (b i ) i k n 1 (a i ) i k. ( ). Suppose A n B, where A is n-indecomposable for ultrafilter U and B is n- indecomposable for ultrafilter V. Then, A n 1 B, so by the inductive hypothesis, t n 1 (A) n 1 t n 1 (B); and thus, ( t n (A) ) ( n 1 n 1 tn (B) ) by Lemma 3.9.2. To n 1 conclude t n (A) n t n (B), it remains to show t n (A) w n t n (B). Let β t n (B). There is a partition (b 0, b 1, b 2 ) of B into (n 1)-indecomposables with b 0 V and t n 1 (b i ) n 1 β (for i {1, 2}) by Lemma 3.15.1. Since β t n (B) we must actually have t n 1 (b i ) = β (for i {1, 2}). There is a matching partition (a 0, a 1, a 2 ) of A with b i n 1 a i for i {0, 1, 2}. Since A is n-indecomposable for U, there is some i 2 such that a i is n-indecomposable for U; for this i, a 0 a i U, so that a 0 a i n a i n 1 A. We also have b 0 b i V, so that b 0 b i n b 0 n 1 B. Thus, b 0 b i n 1 a 0 a i. In what follows we will consider a partition (b 0, b 1 ) of B with b 0 V and t n 1 (b 1 ) = β, and a partition (a 0, a 1 ) of A with a 0 U and b i n 1 a i for i {0, 1}. We complete the proof assuming Lemma 3.18 holds for n 1. Partition a 1 further into (n 1)-indecomposables, (a j) j m, so that a 1 = j m a j. Since b 1 is (n 1)- indecomposable for some ultrafilter and b 1 n 1 j m a j, it follows by Lemma 3.18 that b 1 n 1 a j (for some j m), and thus by the inductive hypothesis, t n 1 (b 1 ) n 1 t n 1 (a j). Since β = t n 1 (b 1 ) n 1 t n 1 (a j) and a j is (n 1)-indecomposable, it follows by Lemma 3.15.2 that β n 1 γ for some γ t n (A). Since β t n (B) was arbitrary, it follows that t n (A) w n t n (B). Proof of Lemma 3.18. We assume Theorem 3.17 holds for n 1 and show the Lemma for n. We prove the following special case: For any a A which is n-indecomposable (for ultrafilter U) and partition (b 0, b 1 ) of B, if a n b 0 b 1 then a n b 0 or a n b 1. The lemma then follows by induction using this special case. Suppose a n b 0 and let (d i ) i k be a partition of b 0 with no matching partition (a i ) i k of a with (a i ) i k n 1 (d i ) i k. To show that a n b 1, consider any partition (e j ) j l of b 1, so that by hypothesis there is a partition (a i ) i k+l such that (d i ) i k n 1 (a i ) i k and (e i ) i l n 1 (a k+i ) i l. Since a is n-indecomposable for ultrafilter U, there is some i k + l with a i U and n-indecomposable for U. We will show that i > k. Suppose

ON THE n-back-and-forth TYPES OF BOOLEAN ALGEBRAS 15 that i k, say a 0 U. Then let d = a 0 a k+1... a k+l U which is n- indecomposable for U. Thus, t n 1 (a 0 ) = t n 1 (U) = t n 1 (d), so that a 0 n 1 d by the inductive hypothesis for Theorem 3.17, and (d, a 1,..., a k ) is a partition of a satisfying (d, a 1..., a k ) n 1 (d i ) i k, contradicting our assumption about the partition (d i ) i k of b 0 ; so, we may assume a k+1 is n-indecomposable for U. Let e = a k+1 a 0... a k, so e is n-indecomposable for U and a k+1 n 1 e. Thus, (e, a k+2,..., a k+l ) is a partition of a with (e j ) j l n 1 (e, a k+2,..., a k+l ). Since the partition (e j ) j l of b 1 was arbitrary, it follows that a n b 1. This completes the proof of Theorem 3.17. 4. An alternative characterization of n-indecomposable We provide an equivalence between our heuristic definition of n-indecomposable types (Definition 3.1) and our definition in terms of accumulation points in the Stone space (Definition 3.5). Along the way we prove some useful properties of n-indecomposables. Our main result here is: Theorem 4.1. For any Boolean algebra A the following are equivalent: (1) A is n-indecomposable for an ultrafilter U (as in Definition 3.5). (2) There is an ultrafilter U such that A n A a for all a U. (3) For every partition (a i ) i k of A, there is some i k with A n A a i (as in Definition 3.1). We first prove (1) (2). Lemma 4.2. For any Boolean algebra A, if A is n-indecomposable for an ultrafilter U, then A n A a for all a U. Proof. If A is n-indecomposable for U, then for all a U, t n (A) n t n (a) and a is also n-indecomposable for U by Lemma 3.12.2. By Theorem 3.17, A n A a for all a U. Lemma 4.3. For all Boolean algebras A and B and for every n ω, if A is (n + 1)- indecomposable for ultrafilter U, then A n+1 B if and only if for every partition (b i ) i k of B into n-indecomposables, there is a partition of (a i ) i k of A into n-indecomposables satisfying (b i ) i k n (a i ) i k. Note that we impose no condition on B as we did in Lemma 3.16. Proof. ( ). This follows by Lemma 3.14. ( ). Suppose A is (n + 1)-indecomposable for ultrafilter U and A n+1 B. Let (b i ) i k be a partition of B into n-indecomposables. Since A is (n + 1)-indecomposable, by Lemma 3.18 we may assume that A n+1 B b 0, which implies A n B b 0. By hypothesis, there is a partition (a i ) i k of A satisfying (b i ) i k n (a i ) i k, where the a i may not be n-indecomposable. Partition each a i into n-indecomposables, (a i,j ) j mi (by Theorem 3.13). We will also assume b i n a i,0 (for each i k) by Lemma 3.18. Since A is (n+1)-indecomposable, by Lemma 4.2 there is some a i,j U so that A n A a i,j. If j = 0 then let a i,0 = a i,0 l k 1 j m l a l,j, so that a i,0 U and b i n a i,0 n a i,0. Otherwise, let a 0,0 = a 0,0 l k 1 j m l a l,j, so that a 0,0 U and b 0 n a 0,0. In

16 KENNETH HARRIS AND ANTONIO MONTALBÁN either case, a i,0 together with a l,0 (for l i) provide a matching partition of A into n-indecomposables. Now we prove (2) (1) of Theorem 4.1. Lemma 4.4. For any Boolean algebra A, if there is an ultrafilter U such that A n A a for all a U, then A is n-indecomposable for U. Proof. Let U be an ultrafilter such that A n A a for all a U. Let a U be n-indecomposable for U (by Lemma 3.12.1). Then A n A a. Since a 1 A, it follows that t n (A) w n t n (a). We will show the following: ( ) For any V U, t n 1 (V ) w t n (a). Let V U and choose pairwise disjoint (n 1)-indecomposable elements b 0 U and b 1 V ; then set b 2 = 1 A (b 0 b 1 ). Since A a n A and a is n-indecomposable for U, by Lemma 4.3 there is a partition of a, (a 0, a 1, a 2 ), into (n 1)-indecomposables for some ultrafilter such that (b 0, b 1, b 2 ) n 1 (a 0, a 1, a 2 ). By Theorem 3.17, we then have t n 1 (U) n 1 t n 1 (a 0 ) and t n 1 (V ) n 1 t n 1 (a 1 ). First, suppose that a 1 U. We have that t n 1 (a 0 ) w t n (a) by condition (ii) of Definition 3.5, as a 0 is (n 1)-indecomposable for some ultrafilter W U. Thus, t n 1 (U) w t n (a). Since a 1 a, by Lemma 3.12.2 a 1 is n-indecomposable for U, and so t n 1 (V ) n 1 t n 1 (a 1 ) n 1 t n 1 (U) w t n (a). Now, suppose that a 1 U. Then it follows that t n 1 (V ) n 1 t n 1 (a 1 ) w t n 1 (a) by condition (ii) of Definition 3.5, as a 1 is (n 1)-indecomposable for some ultrafilter W U. Thus, ( ) holds. It follows from ( ) that t n (a) w n t n (A), and since t n (A) w n t n (a), from Lemma 3.8.c it follows that t n (A) n t n (a) n t n (U), and thus condition (i) of Definition 3.5 holds. Condition (ii) of Definition 3.5 follows straightforwardly from ( ) and t n (A) n t n (a). Since (2) = (3) follows from the property Uf.a of ultrafilters, there only remains showing (3) = (2) to complete Theorem 4.1. Let A satisfy the conditions in (3), so that for every partition (a i ) i k of A, there is some i k with A n A a i. Say a A is strongly n-equivalent to A if A a n A and for any partition (c i ) i k of a, there is some i with A n A c i. Note that (3) says that 1 A is strongly n-equivalent to A. We will show that for every a A which is strongly n-equivalent to A, there is an ultrafilter U a all of whose members are strongly n-equivalent to A. This ultrafilter will also be sufficient for (2). The proof is by induction on n, where the basis case n = 0 is trivial, since every nonzero element of A is strongly 0-equivalent to A. Assume that if a is strongly (n 1)-equivalent to A, then there is an ultrafilter V a all of whose members are strongly (n 1)-equivalent to A. Claim 1. If a A is strongly n-equivalent to A, then for every partition (a i ) i k of a, there is an i with a i strongly n-equivalent to A. Proof. Suppose a A is strongly n-equivalent to A and (a i ) i k is a counterexample to the claim. Then for each i, there is a partition (b i,j ) j mi of a i for which there is no

ON THE n-back-and-forth TYPES OF BOOLEAN ALGEBRAS 17 j m i with A n A b i,j. Now, (b i,j ) i k,j mi is a partition of a with no i k, j m i satisfying A n A b i,j, contradicting our assumption about a. Claim 2. If a is strongly n-equivalent to A and a b, then A n A b. Proof. Since a is strongly n-equivalent to A, it is also strongly (n 1)-equivalent to A, so by the induction hypothesis there is an ultrafilter V a all of whose members are strongly (n 1)-equivalent to A. It follows that a is (n 1)-indecomposable for V. We first show that A n A b. Let (b i ) i k be a partition of b; so, (a b i ) i k is a partition of a. Thus, for some i k, a b i V, and in particular, A a b i n 1 A a n 1 A. We may suppose i = 0; then let c j = b j and c 0 = 1 A (b 1... b k ). Thus, (c i ) i k is a partition of A and c i n 1 b i for i > 0, and for i = 0 we have c 0 V since a b 0 V and V is an ultrafilter. So, (c i ) i k is a partition of A with (c i ) i k n 1 (b i ) i k. Thus, A n A b. We now show that A b n A. Now, let (c i ) i k be a partition of 1 A. Since A a n A, there is a partition (a i ) i k of a with A c i n A a i. We may also suppose that a 0 V since a V and V is an ultrafilter. Let b i = a i for i > 0 and b 0 = b (a 0... a k ), so that (b i ) i k is a partition of b. We still have c i n 1 b i for i > 0. But, A a 0 n 1 A b 0 since a 0 V and a 0 b 0 implies b 0 V. Thus, (b i ) i k is a partition of b with (c i ) i k n 1 (b i ) i k, so that A b n A. Claim 3. If a is strongly n-equivalent to A and b a, then b is strongly n-equivalent to A. Proof. Let a be strongly n-equivalent to A and b a. Consider a partition (c i ) i<l of b. Then (c i a) i<l is a partition of a, and hence some c i a is strongly n-equivalent to A by the first claim. But then, by the second claim, A c i n A. Since the partition (c i ) i<l of b was arbitrary, it follows that b is strongly n-equivalent to A. Let a be strongly n-equivalent to A and suppose that for every ultrafilter U a, there is an element b U U which is not strongly n-equivalent to A. Since Ult(A) is compact, so is the clopen set O a = { U Ult(A) : U a }, and so there is a finite set {b U0,..., b Ul } such that b U0... b Ul = a. (This argument is as given in the proof of Theorem 3.13.) Then there is a partition (b i,j ) i l,j mi of a with b Ui = j m i b i,j. Since none of the b Ui are strongly n-equivalent to A, by the third claim none of the b i,j can be strongly n-equivalent to A, contradicting the hypothesis that a is strongly n-equivalent to A. Therefore, there is an ultrafilter U a all of whose members are strongly n-equivalent to A. 5. Realizable n-bf-indecomposable types. In this section we characterize the n -antichains of ABF n which are realized in some (n + 1)-indecomposable Boolean algebra. Definition 5.1. We say that α ABF n is realizable if there exists an (n + 1)- indecomposable Boolean algebra A such that t n+1 (A) = α. We use BF n to denote the set of realizable antichains of ABF n 1. Theorem 5.3 below gives a simple combinatorial property that characterizes the realizable n-bf-types. Moreover, for every realizable n-bf-type, the proof explicitly builds a (computable) Boolean algebra of that type.